归并排序求逆序数

思路：假设前半部分（A[left]到A[mid]）与后半部分（A[mid+1]到A[right]）都是从小到大排好序的，那么如果A[left]>A[mid+1]，则A[mid+1]与前半部分的逆序数目是mid-left+1

int merge(int *A, int left, int mid, int right){
int *temp = new int[right-left+1];
int num = 0;
int i = left;
int j = mid+1;
int index = 0;

while(i<=mid && j<=right){
if(A[i]>A[j]){
num+=mid-i+1;
temp[index]=A[j];
j++;
}
else{
temp[index]=A[i];
i++;
}
index++;
}
if(i<=mid)
for(;i<=mid;i++){
temp[index]=A[i];
index++;
}
if(j<=right)
for(;j<=right;j++){
temp[index]=A[j];
index++;
}

for(int k=0; k<index; k++)
A[left+k]=temp[k];

delete []temp;
return num;
}

int inversion(int *A, int left, int right){
if(left>=right)  return 0;

int mid=(left+right)/2;
int num1=inversion(A,left,mid);
int num2=inversion(A, mid+1, right);
return num1+num2+merge(A,left,mid,right);
}

int main(){
int A[5]={9,8,7,6,5};
cout << inversion(A,0,4);
return 0;
}